9. Ray Optics And Optical Instruments

Big idea: draw where light actually travels before deciding where an image appears.

This chapter uses one stated Cartesian convention and distinguishes image position, orientation and angular magnification.

Early telescopes and microscopes combined lenses long before complete wave theories of light. Later optical design used both ray geometry for image formation and wave theory for resolution and diffraction limits.

A camera, a mirror and a microscope all form images by redirecting light.

Ray optics represents light by directed lines normal to wavefronts in an isotropic medium. It works well when relevant apertures and obstacles are much larger than the wavelength. It does not explain diffraction fringes; those need a wave model.

A real image is formed where actual outgoing rays converge and can be received on a screen. A virtual image is where backward extensions of diverging rays appear to meet. An eye can see a virtual image, but placing an ordinary screen at its apparent position does not capture it.

For every calculation, draw a small diagram, select a coordinate direction, assign signed distances, solve, and interpret the result physically. “Negative image distance” has no universal meaning without a specified convention and optical element.

A flat mirror appears to place an object behind its surface. Can a screen behind the opaque mirror receive that apparent image?

Reflected rays remain in front of the mirror. Only their backward extensions meet behind it.

Result: No. It is a virtual image, although an observer in front can see it.

1. When is geometric optics a useful approximation?
2. What distinguishes real and virtual images?

Answers

1. When wavelength is small compared with the important dimensions and diffraction is negligible.
2. Actual rays converge for a real image; their backward extensions meet for a virtual image.

For a spherical mirror, the axis and sign convention organize both the drawing and the algebra.

Take the mirror pole as origin and incident light travelling from left to right as positive x. Measure axial distances from the pole: objects in front have negative u. Heights above the axis are positive. A concave mirror facing the object has its centre and principal focus on the negative side, so its R and f are negative. A convex mirror has positive R and f.

For paraxial rays, a spherical mirror has $f=R/2$. A concave mirror converges rays initially parallel to the axis at its principal focus. A convex mirror makes reflected parallel rays diverge as if from its virtual focus. Large-angle rays do not all meet at the paraxial focus: spherical aberration limits the simple formula.

Using the small-angle approximation in reflection geometry gives $1/v+1/u=2/R=1/f$. Transverse magnification is $m=h_i/h_o=-v/u$. First solve signed distances; then use the sign of m for orientation and its magnitude for relative size.

• A ray parallel to the principal axis reflects through the focus, or appears to come from it.
• A ray directed through the centre of curvature retraces its path because it strikes normally.
• A ray at the pole reflects with equal angles to the principal axis.

An object is $48\,\mathrm{cm}$ in front of a concave mirror with $f=-16\,\mathrm{cm}$. Find image position and magnification.

Set $u=-48\,\mathrm{cm}$ and use $1/v=1/f-1/u$. Then use $m=-v/u$.

Result: $v=-24\,\mathrm{cm}$, $m=-0.5$. The image is real, inverted and diminished, in front of the mirror.

To see the paraxial origin of the mirror equation, take a ray striking at a small height h. Express the incident-ray, reflected-ray and normal slopes by their signed axial distances. Equality of incidence and reflection angles gives the small-angle relation $h/u+h/v=2h/R$. Cancelling nonzero h gives $1/u+1/v=2/R$. The magnification sign follows independently by comparing the object and image triangles about the pole.

Moving an object inside a concave mirror’s focal distance produces a virtual upright enlarged image behind it. A convex mirror with an ordinary real object produces a virtual upright diminished image. At the focus, outgoing paraxial rays are parallel and the ideal image is at infinity.

1. Which approximation supports f=R/2 for a spherical mirror?
2. What does negative magnification mean?

Answers

1. Paraxial or small-angle rays.
2. The image is inverted relative to the chosen positive object height.

At an interface, frequency stays fixed while speed and wavelength can change.

The refractive index relative to vacuum is $n=c/v$. For isotropic media the incident ray, refracted ray and normal lie in one plane. Snell’s law is $n_1\sin i=n_2\sin r$, with angles measured from the normal. Reflection can occur at the same boundary as transmission.

Entering a higher-index medium bends an oblique ray towards the normal. Entering a lower-index medium bends it away when transmission is possible. At normal incidence there is no angular bending, although speed and wavelength may change. Dispersion means that n depends on wavelength.

Light enters glass of index $1.50$ from air at incidence $45^\circ$. Find the refracted angle, treating air index as one.

$r=\sin^{-1}(\sin45^\circ/1.50)$.

Result: $r=28.1255^\circ$.

For near-normal viewing across a plane interface, apparent depth equals real depth times $n_{\rm observer}/n_{\rm object}$. Thus an object in water seen from air appears shallower. This expression is paraxial; it is not an exact rule for all viewing angles.

1. Does frequency change at a stationary glass–air boundary?
2. When can refraction change speed without changing the ray direction?

Answers

1. No.
2. At normal incidence.

Transmission can disappear when a ray tries to leave a higher-index medium at too large an angle.

Total internal reflection requires both $n_1>n_2$ and an incident angle greater than the critical angle. At the critical angle the ideal refracted ray grazes the surface, so $\sin i_c=n_2/n_1$. Below this angle part of the light can transmit; above it, an ideal transparent interface reflects the incident power in the geometrical model.

Find the critical angle from a material with $n_1=1.60$ into air of $n_2=1.00$.

$i_c=\sin^{-1}(1.00/1.60)$.

Result: $i_c=38.6822^\circ$.

A clear container, a directed beam and a suitable curved entry surface can demonstrate this transition while changing incidence at the final plane boundary. Describe the angle at the internal exit surface, not the angle at the entry face.

Prisms can redirect beams using internal reflection. An optical fibre guides suitable rays within a higher-index core surrounded by lower-index cladding. Each reflection must satisfy the core–cladding condition; not every possible entering ray is guided. Bending a fibre too sharply can increase leakage.

A mirage results from refraction through a gradient of refractive index in air, often associated with a temperature gradient. It should not be explained as reflection from a real pool of water. Sparkle in a cut transparent solid depends on refraction, dispersion and repeated internal reflections, not on critical angle alone.

1. What happens at exactly the critical angle in ray optics?
2. Why does fibre cladding need a lower index than the core?

Answers

1. The transmitted ray is tangent to the interface.
2. It allows total internal reflection for suitable internal incidence angles.

A lens combines refraction at two curved surfaces; keep the signs tied to the same axis throughout.

For paraxial rays going from index $n_1$ into $n_2$, $n_2/v-n_1/u=(n_2-n_1)/R$. Measure u, v and R from the surface pole along the positive incident-light direction. The radius is positive when its centre lies on that positive side. The relation follows by combining Snell’s small-angle form with the slopes of the ray and surface normal.

The spherical-surface relation handles both real and virtual intermediate images. When two interfaces are combined, the first image acts as the object for the second; the sign of that intermediate object follows its location, not a memorized label.

For the first air-to-glass surface, write $n/v_1-1/u=(n-1)/R_1$. For the glass-to-air surface, write $1/v-n/u_2=(1-n)/R_2$. A thin lens has $u_2\approx v_1$ with this common axial convention. Adding cancels the intermediate-distance terms and gives $1/v-1/u=(n-1)(1/R_1-1/R_2)$. For a distant object, $1/u\to0$ and the image lies at f, identifying the lens-maker expression.

Neglecting lens thickness and combining the two surface equations for a lens in air gives $1/v-1/u=1/f$. The lens-maker relation is $1/f=(n-1)(1/R_1-1/R_2)$. For a lens in a surrounding medium, replace n by the relative index of lens to medium. A lens that converges in air need not converge in every surrounding material.

Transverse magnification of a thin lens is $m=v/u$. A converging lens has positive f; a diverging lens has negative f under this convention. Unlike the mirror expression, the thin-lens formula contains a minus sign before $1/u$. Do not exchange the two equations.

An object is $36\,\mathrm{cm}$ to the left of a converging lens with $f=12\,\mathrm{cm}$. Find image distance and magnification.

$u=-36\,\mathrm{cm}$. Use $1/v=1/f+1/u$ and $m=v/u$.

Result: $v=18\,\mathrm{cm}$, $m=-0.5$. The image is real, on the right, inverted and half the object size.

Lens power is $P=1/f$ with f in metres; its unit is dioptre, $\mathrm{m^{-1}}$. Powers of thin lenses in contact add: $P_{\rm eq}=P_1+P_2$. A positive result is converging. For separated lenses, first form the image from one lens, then locate it relative to the next lens; simple power addition no longer applies.

A $+5.0\,\mathrm D$ lens is in contact with a $-2.0\,\mathrm D$ lens. Find equivalent power and focal length.

Add the signed powers, then invert the result using metres.

Result: $P=3\,\mathrm D$, $f=0.3333\,\mathrm m$.

The earlier $0.12\,\mathrm m$ lens has power $8.3333\,\mathrm D$. This is not the same number obtained by inserting its focal length in centimetres.

1. What must be true before lens powers can simply be added?
2. What sign does a diverging lens focal length have?
3. Where does a positive R point?

Answers

1. Thin lenses must be in contact in the same surrounding medium, within the paraxial model.
2. Negative.
3. From the surface pole towards a centre of curvature on the positive incident-light axis.

A prism changes direction at two faces, and each refraction must be accounted for.

Let prism angle be A, internal refraction angles $r_1,r_2$, incidence i and emergence e. Geometry gives $r_1+r_2=A$ and deviation $\delta=i+e-A$. The two boundary refractions obey Snell’s law separately.

At minimum deviation in the symmetric air–prism–air path, $i=e$ and $r_1=r_2=A/2$. Therefore $n=\sin[(A+\delta_m)/2]/\sin(A/2)$. For a thin prism and small angles in radians, $\delta\approx(n-1)A$. The latter is an approximation, not a replacement for all prism calculations.

A prism has $A=60^\circ$ and minimum deviation $36^\circ$ in air. Find its refractive index for that wavelength.

Use the symmetric minimum-deviation formula with half-angles.

Result: $n=1.4863$.

Dispersion separates colours because refractive index depends on wavelength. In ordinary visible dispersion, violet is generally deviated more than red by the same prism. A prism does not create the frequencies in white light; it sends existing components in different directions.

1. What relation holds between entry and exit angles at minimum deviation?
2. Why does a prism separate colours?

Answers

1. They are equal for the symmetric path in the same outside medium.
2. Its refractive index, and hence deviation, depends on wavelength.

Optical instruments make objects subtend a larger angle at the eye.

A positive lens used with an object inside its focal distance gives an upright virtual image. Angular magnification compares the viewing angle with the angle of the unaided object at the conventional near distance D. For a relaxed eye with final image at infinity, $M=D/f$. With the eye close to the lens and final image at the near point, $M=1+D/f$. These are angular, not transverse image-size ratios.

The short-focal-length objective makes a real enlarged inverted intermediate image. The eyepiece acts as a magnifier of that image. For a sufficiently long tube, objective magnification magnitude is approximately $L/f_o$, with L representing the relevant optical tube length in this approximation. For normal adjustment, total magnitude is approximately $(L/f_o)(D/f_e)$. For near-point viewing, replace the eyepiece factor by $1+D/f_e$.

Using the approximate normal-adjustment model, take $L=16\,\mathrm{cm}$, $f_o=0.80\,\mathrm{cm}$, $f_e=4.0\,\mathrm{cm}$ and $D=25\,\mathrm{cm}$. Find magnifying power magnitude.

Multiply the objective and eyepiece magnifications. Use the same length unit throughout.

Result: $|M|=125$. The final image is inverted relative to the original object.

A distant object supplies nearly parallel rays. A long-focal-length objective forms a real image near its focal plane. The eyepiece views that image. In normal adjustment the final image is at infinity, separation is approximately $f_o+f_e$, and signed angular magnification is $M=-f_o/f_e$. The minus sign denotes inversion.

An astronomical telescope has $f_o=90\,\mathrm{cm}$ and $f_e=3.0\,\mathrm{cm}$. Find normal-adjustment magnification and lens separation.

Use the focal-length ratio and their sum.

Result: $M=-30$ and separation $93\,\mathrm{cm}$.

A larger objective aperture gathers more light and can improve diffraction-limited resolution. Magnification alone cannot reveal detail that the objective fails to resolve. Reflecting telescopes use a curved mirror objective and avoid the objective’s chromatic refraction; secondary-mirror arrangements fold the optical path. Real instruments also face aberration, alignment and atmospheric limits.

1. What is the intermediate image in a compound microscope?
2. Why is telescope objective focal length usually much larger than eyepiece focal length?
3. Does increasing magnification always improve resolved detail?

Answers

1. A real enlarged inverted image formed by the objective.
2. The ratio gives a large angular magnification in normal adjustment.
3. No. Resolution is constrained by aperture, wavelength and other optical limits.

• Reflection: $i=r$. Mirror: $1/v+1/u=1/f$, $m=-v/u$.
• Refraction: $n_1\sin i=n_2\sin r$. TIR requires higher-to-lower index and $i>i_c$.
• Spherical interface: $n_2/v-n_1/u=(n_2-n_1)/R$.
• Thin lens: $1/v-1/u=1/f$, $m=v/u$, $P=1/f$ with f in metres.
• Prism: $\delta=i+e-A$ and minimum-deviation symmetry.
• Normal telescope: $M=-f_o/f_e$; distinguish angular magnification from image-height magnification.

• Image signs are meaningful only with a declared convention.
• An enlarged image can still contain no additional resolved detail.
• Thin-lens, paraxial and near-normal formulas have stated limits; do not silently extend them.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

9.1 State the Cartesian sign convention used here.

9.2 Find v for a concave mirror with $f=-12\,\mathrm{cm}$ and $u=-36\,\mathrm{cm}$.

9.3 For that mirror image, find m.

9.4 A convex mirror has $f=20\,\mathrm{cm}$ and $u=-60\,\mathrm{cm}$. Find v.

9.5 Why is f=R/2 an approximation for a spherical mirror?

9.6 Find the speed of light in a medium of index $1.25$ using $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$.

9.7 Why can a normally incident ray change wavelength without bending?

9.8 A $0.24\,\mathrm m$ depth in index $1.50$ is viewed nearly normally from air. Find apparent depth.

9.9 Find critical angle from index $1.40$ into air.

9.10 List the two requirements for total internal reflection.

9.11 Does every ray entering an optical fibre remain guided?

9.12 A positive lens has $f=15\,\mathrm{cm}$, $u=-45\,\mathrm{cm}$. Find v and m.

9.13 A diverging lens has $f=-18\,\mathrm{cm}$, $u=-36\,\mathrm{cm}$. Find v.

9.14 Find power for $f=0.25\,\mathrm m$.

9.15 Find equivalent power of $+6\,\mathrm D$ and $-1.5\,\mathrm D$ thin lenses in contact.

9.16 Why cannot powers of widely separated lenses simply be added?

9.17 For $n=1.50$, $R_1=0.20\,\mathrm m$, $R_2=-0.20\,\mathrm m$, find thin-lens power in air.

9.18 When does a converging lens make an upright virtual image of a real object?

9.19 How can an eye see a virtual image if no rays meet there?

9.20 A prism has $A=50^\circ$, $i=40^\circ$, $e=45^\circ$. Find deviation.

9.21 State the equalities at minimum prism deviation.

9.22 Explain normal visible dispersion in a prism.

9.23 Find normal-adjustment magnifier power for $D=25\,\mathrm{cm}$, $f=5\,\mathrm{cm}$.

9.24 For the same magnifier, find near-point angular magnification in the close-eye approximation.

9.25 Distinguish transverse and angular magnification.

9.26 A normal-adjustment telescope has $f_o=120\,\mathrm{cm}$, $f_e=4\,\mathrm{cm}$. Find signed M and separation.

9.27 Why is a larger telescope objective useful even at unchanged magnification?

9.28 Use the approximate microscope formula for $L=18\,\mathrm{cm}$, $f_o=1.2\,\mathrm{cm}$, $D=25\,\mathrm{cm}$, $f_e=5\,\mathrm{cm}$.

9.29 What advantage does a reflecting objective have concerning chromatic aberration?

9.30 A plane mirror rotates through a small angle θ with fixed incident ray. How much does the reflected direction turn?

9.31 Why should a ray diagram and signed formula agree?

Answers

9.1 Origin at pole or optical centre; incident-light direction is positive; heights above the axis are positive.

9.2 $v=-18\,\mathrm{cm}$.

9.3 $m=-0.5$: inverted and diminished.

9.4 $v=15\,\mathrm{cm}$: virtual behind the mirror.

9.5 It assumes paraxial rays; large-angle rays suffer spherical aberration.

9.6 $240000000\,\mathrm{m\,s^{-1}}$.

9.7 Frequency is fixed while speed changes; the ray still follows the normal.

9.8 $0.16\,\mathrm m$.

9.9 $45.5847^\circ$.

9.10 Travel towards lower index, with incidence above the critical angle.

9.11 No. It must meet the internal guidance condition and avoid excessive leakage from bending or defects.

9.12 $v=22.5\,\mathrm{cm}$, $m=-0.5$.

9.13 $v=-12\,\mathrm{cm}$, a virtual image on the object side.

9.14 $4\,\mathrm D$.

9.15 $4.5\,\mathrm D$.

9.16 The first image location determines the signed object distance for the second; separation changes the mapping.

9.17 $5\,\mathrm D$.

9.18 When the object lies between the lens and its principal focus.

9.19 Rays entering the eye appear to diverge from that position; the eye forms its own real retinal image.

9.20 $35^\circ$.

9.21 i=e and the internal angles both equal A/2 for the symmetric path.

9.22 The index depends on wavelength; ordinary glass generally deviates violet more than red.

9.23 $5$.

9.24 $6$.

9.25 Transverse magnification compares heights; angular magnification compares angles subtended at the observer.

9.26 $M=-30$ and $124\,\mathrm{cm}$.

9.27 It gathers more light and can improve diffraction-limited resolving power.

9.28 $|M|=75$.

9.29 Reflection does not use the wavelength-dependent refraction of a lens objective.

9.30 Through 2θ, in the same plane of rotation.

9.31 They describe the same paraxial geometry; disagreement signals a sign, ray-rule or approximation mistake.